CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set E

Question. Find the A.P. whose n^th term is given by : (a) 9 – 5n (b) –n + 6 (c) 2n + 7
Solution. (a) 4, – 1, –6, –11, ..... (b) 5, 4, 3, 2, ....... (c) 9, 11, 13, 15, ......

Question. If the 10^th term of an A.P. is 52 and 17^th term is 20 more than the 13^th term, find the A.P.
Solution. 7, 12, 17, 22.......

Question. The angles of a triangle are in A.P. If the greatest angle equals the sum of the other two, find the angles.
Solution. 30°, 60°, 90°

Question. Three numbers are in A.P. If the sum of these numbers is 27 and the product is 648, find the numbers.
Solution. 6, 9, 12

Question. Find the three numbers in A.P. such that their sum is 24 and the sum of their squares is 194.
Solution. 7, 8 and 9

Question. How many terms of the sequence –35, –28, –21, ...... should be taken so that their sum is zero?
Solution. 11

Question. How many terms of the A.P. 63, 60, 57,..... must be taken so that their sum is 693? 
Solution. 21 or 22

Question. How many numbers of three digits are exactly divisible by 11?
Solution. 81

Question. Two A.P.’s have the same common difference. The difference between their 100th terms is 111222333. What is the difference between their millionth terms?
Solution. 111222333

Question. (a) For an A.P., find Sn if given a_n = 3 – 5n.
(b) For an A.P., find an if given S_n = 2n² + 5n.
Solution. (a) n/2 (1 –5n) (b) 4n + 3

Question. Find the sum of first n natural numbers.
Solution. n(n + 1)/2

Question. For what value of n, the n^th terms of the two A.P.’s are same? 3, 10, 17, ...... and 63, 65, 67, .....
Solution. n = 13

Question. Find the sum of first 25 odd natural numbers.
Solution. 625

Question. Find the number of integers between 200 and 500 which are divisible by 7.
Solution. 43

Question. Find the value of k so that k – 3, 4k – 11 and 3k – 7 are three consecutive terms of an A.P.
Solution. k = 3

Question. The 4^th term of an A.P. is three times the first and the 7^th term exceeds twice the third term by 1. Find the first term and the common difference.
Solution. a = 3, d = 2

Question. Find the sum of n terms of an A.P. whose nth term is given by an = 5 – 6n.
Solution. 2n –3n²

Question45. In an A.P., an = 116, a = 2 and d = 6. Find Sn.
Solution. 1180

Question. If the sum of the first 6 terms of an A.P. is 36 and the sum of the first 21 terms is 441, find the first term and the common difference and hence find the sum of n terms of this A.P.
Solution. a = 1, d = 2; Sn = n²

Question. Find the A.P. whose third term is 16 and the seventh term exceeds its fif^th term by 12.
Solution. 4, 10, 16, ......

Question. The 8thterm of an A.P is 0 prove that its38t h term is triple its18th term.
Solution. -800

Question. Which term of the A.P 12,7,2,-3……is -98
Solution. 23rdterm

Question. between 250 and 1000 which are exactly divisible by 3.
Solution. 156375

Question. Find the value of x so that 3x + 2, 7x – 1 and 6x + 6 are three consecutive terms of an A.P.
Solution. x = 2

Question. A sum of Rs. 1000 is invested at 8% simple interest per annum. Calculate the interest at the end of 1, 2, 3, ......years. Is the sequence of interest an A.P.? Find the interest at the end of 30 years.
Solution. Rs. 80, Rs. 160, Rs. 240, and so on. Yes; Rs. 2400

Question. between 100 and 1000 which are multiples of 5.
Solution. 98450

Question. For what value of n is the n^th term of the following A.P.’s the same? 1, 7, 13, 19, ..... and 69, 68, 67............... 
Solution. No value of n

Question. Find the middle term of an A.P. with 17 terms whose 5^th term is 23 and the common difference is –2.
Solution. 15

Question. For what value of n, the n^th terms of the two A.P.’s are same? 2, –3, –8, –13, ...... and –26, –27, –28, –29, .......
Solution. n = 8

Question. Which term of the A.P. 3, 10, 17, .... will be 84 more than its 13^th term? 
Solution. 25^th

Question. Find the sum of first 20 terms of an A.P., in which 3rd term is 7 and 7^th term is two more than thrice of its 3^rd term.
Solution. 740

Question. between 50 and 500 which are divisible by 7.
Solution. 17696

Question. Find the 8^th term from the end of the A.P. 7, 10, 13, ..... , 184. 
Solution. 163

Question. The sums of n-terms of two A.P.’s are in the ratio 7n + 1 : 4n + 27. Find the ratio of their 11^th terms.
Solution. 4 : 3

Question. The n^th term of an A.P. is given by tn = 4n – 5. Find the sum of the first 25 terms of the A.P.
Solution. 1175

Question. Two cars start together in the same direction from the same place. The first goes with a uniform speed of 10 km/hr. The second goes with a speed of 8 km/hr in the first hour and increases the speed by 1/2 km/hr in each succeeding hour. After how many hours will the second car overtake the first if both cars go nonstop?
Solution. 9 hours

Question. The ages of the students in a class are in A.P. whose common difference is 4 months. If the youngest student is 8 years old and the sum of the ages of all the students is 168 years, find the number of students in the class.
Solution. 16

Question. Find the sum of first 30 even natural numbers.
Solution. 930

Question. The 10^th term of an A.P. is 29 and and sum of the first 20 terms is 610. Find the sum of the first 30 terms.
Solution. 1365

Question. The sum of the first 15 terms of an A.P. is 105 and the sum of next 15 terms is 780. Find the first three terms of the A.P.
Solution. –14, –11, –8

Question. A person borrows Rs. 4500 and promises to pay back (without any interest) in 30 instalments each of value Rs. 10 more than last (preceding one). Find the first and the last instalments.
Solution. Rs. 5, Rs. 295

Question. The 9^th term of an A.P. is equal to 7 times the 2^nd term and 12^th term exceeds 5 times the 3rd term by 2. Find the first term and the common difference.
Solution. a = 1, d = 6

Question. Find the 8^th term of an A.P. whose 15th term is 47 and the common difference is 4.
Solution. 19

Question. How many terms of an A.P. 1, 4, 7, ..... are needed to give the sum 1335?
Solution. 30

Question. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first ten terms of the A.P.
Solution. 32.5 or 27.5

Question. How many terms of an A.P -6 , 11/2 , -5 ..... are needed to give the sum –25? Explain the double answer.
Solution. 5 or 20

Question. If the sum of the first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.
Solution. It is given that
S₆ = 36 and S₁₆ = 256
Let a be the first term and d be the common difference of an AP.
We know that sum of n terms of an AP,
S_n =n/2[2a + (n – 1)d]
Now S6 = 36 (Given)
⇒ 6/2[2a + (6 – 1)d] = 36
⇒ 3 [2a + (6 – 1)d = 36
⇒ 2a + 5d = 12
Also, S₁₆ = 256
⇒16/2[2a + (16 – 1)d] = 256
⇒ 8[2a + 15d] = 256
⇒ 2a + 15d = 32 ...(ii)
Subtracting equation (i) from equation (ii), we get
(2a + 15d) – (2a + 5d) = 32 – 12
⇒ 2a + 15 – 2a – 5d = 20
⇒ 10d = 20
⇒ d = 2
Putting the value of d in equation (i), we get
2a + 5d = 12
2a + 5(2) = 12
⇒ 2a + 10 = 12
⇒ 2a = 2 ⇒ a = 1
We have to find sum of first 10 terms, S₁₀ = ?
S₁₀ =10/2[2a + (10 – 1)d]
= 5 [2(1) + 9(2)]
= 5 [2 + 18]
= 5 × 20
S₁₀ = 100
Hence, the required sum of the first 10 terms is 100.

Question. The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is -30 and the common difference is 8. Find n.
Solution. It is given that
first term of first AP, a = 8
and common difference of first AP, d = 20
Let n be the number of terms in first AP.
We know that sum of first n terms of an AP,
S_n =n/2[2a + (n – 1)d]
=n/2[2 × 8 + (n – 1)20]
=n/2[16 + 20n – 20]
=n/2[20n – 4] = n[10n – 2]
⇒ S_n = n[10n – 2] ...(i)
Now, first term of second AP (a’) = –30
Common difference of secod (d’) = 8
∴ Sum of first 2n terms of second AP
S_2n =2n/2 [2a’ + (2n – 1)d’]
= n[2(–30) + (2n –1)8]
= n[–60 + 16n – 8]
S_2n = n[16n – 68] ...(ii)
By given condition,
Sum of first n terms of first AP
= sum of first 2_n terms of second AP
⇒ Sn = S_2n
Using equation (i) and equation (ii), we get
⇒ n(10n – 2) = n(16n – 68)
⇒ 10n – 2 – 16n + 68 = 0
⇒ –6n + 66 = 0
⇒ –6(n – 11) = 0
⇒ n = 11
Hence, the required value of n is 11.

Question. If Sn denotes the sum of first n terms of an AP, prove that S₁₂ = 3(S₈ – S₄)
Solution. We know that sum of n terms of an AP,
S_n =n/2[2a + ( n – 1) d]
S₄ =4/2 [2a + (4 – 1)d]
= 2 [2a +3d] = 4a + 6d
S₈ =8/2[2a + (8 – 1)d]
= 4 [2a +7d] = 8a + 28d
(S8 – S4) = (8a + 28d) – (4a +6d)
= 8a + 28d – 4a – 6d
= 4a + 22d
S₁₂ =12/2 [2a + (12 – 1)d]
= 6 [2a +11d] = 12a + 66d
= 3 [4a + 22d]
S₁₂ = 3 [S8 – S4] [using equation (i)]
Hence proved.

Question. The 14^th term of an AP is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.
Solution. a₁₄ = 2a₈
⇒ a + 13d = 2(a + 7d) ⇒ a = –d
a₆ = –8
⇒ a + 5d = –8
solving to get a = 2, d = –2
S₂₀ = 10(2a + 19d)
= 10(4 – 38)
= –340

Question. The sum of four consecutive number in AP is 32 and the ratio of the product of the first and last terms to the product of two middle term is 7 : 15. Find the numbers.
Solution. Let ‘a' be the first term and 'd’ be the common
difference to the AP. then,
a – 3d, a – d, a + d, a + 3d,
are four consecutive terms of the AP.
As per the question,
(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32, or a = 8 ...(i)
and (a - 3d)(a+3d)/(a - d)(a + d) = 7/15
⇒ a2 - 9d² / a² - d² = 7/15
⇒ 15a² – 135d² = 7a² – 7d²
⇒ 8a2 = 128 d²
⇒ sing (i), we have:
8 × 82 = 128 d²
⇒ d² = 4, or d = ± 2
Thus, the four numbers are 2, 6, 10 and 14

Question. Find the sum of the first 40 positive integers which give a remainder 1 when divided by 6.
Solution. The numbers which when divided by 6 gives 1 as remainder are :
7, 13, 19, 25, 31, 37, ...
They form an A.P., as their common difference
is the same, d = 6
Here: first term, a = 7, common difference, d = 6
Sum of first 40 positive integers:
S₄₀ =n/2[2a + (n – 1)d]
=40/2[2 × 7 + (40 – 1) × 6]
= 20[14 + 39 × 6]
= 20 × 248
= 4,960
Hence, the sum of the first 40 positive integers is 4,960.

Question. Divide 56 in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.
Solution. Let the four parts of the A.P. are (a – 3d), (a – d), (a + d), (a + 3d)
Now, a – 3d + a – d + a + d + a + 3d = 56
⇒ 4a = 56
⇒ a = 14
According to question,
(a−3d)(a+3d)/(a-d)(a+d) = 5/6
⇒ (14 - 3d)(14 + 3d) / (14 - d) (14 + d) = 5/6 [Putting a = 14]
⇒ 196 - 9d2 / 196 - d² = 5/6
⇒ 1176 – 54d² = 980 – 5d²
⇒ 49d² = 196
⇒ d² = 4 ⇒ d = ±2
when,
a = 14 and d = 2
The 4 parts are (a – 3d), (a – d), (a + d), (a + 3d), i.e., 8, 12, 16, 20.
when, a = 14 and d = –2
The 4 parts are 20, 16, 12 and 8.

Question. The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference. 
Solution. Let a be the first term, d be the common difference and n be the number of terms.
It is given that
first term, a = –5
Last term, L = 45
We know that, if the last term of an AP is
known, then the sum of n terms of an AP is
Sn =n /2 (a +l )
120 =n/2(-5 + 45)
120 × 2 = 40 × n
⇒ n = 6
L = a + (n – 1)d
⇒ 45 = –5 + (6 – 1)d
⇒ 50 = 5d ⇒ d = 10
Hence, number of terms = 6
and common difference = 10

Question. Find the sum of first 24 terms of an A.P. whose n^th term is given by a_n = 3 + 2_n.
Solution. Given, nth term of an A.P., an = 3 + 2_n
First term of A.P., a₁ = 3 + 2 × 1 = 5
Second term of A.P., a₂ = 3 + 2 × 2 = 7
third term of A.P., a₃ = 3 + 2 × 3 = 9
and 24th term of A.P., a₂₄ = 3 + 2 × 24 = 51
common difference of A.P., d = a₂ – a₁ = a₃ – a₂
= 7 – 5 = 9 – 7 = 2
Sum of 24 terms, S24 = n /2 [2a+(n - 1)d]
Here, a = 5, n = 24, d = 2
S₂₄ =24/2[2 x 5 + (24 - 1) x2]
= 12 [10 + 46]
= 12 × 56 = 672
Hence the sum of first 24 terms of an A.P. is 676.

Question. If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero.
Solution. Sm = Sn
⇒ = m/2[2a + (n − 1)d]
= n/2[2a + (n − 1)d]
⇒ 2a(m − n) + d(m² − m − n² + n) = 0
⇒ (m − n)[2a + (m + n − 1)d] = 0
or Sm + n = 0

Question. For what value of n are the nth terms of two A.P.’s 63, 65, 67, .... and 3, 10, 17, ..... equal ?
Solution.Given, APs are 63, 65, 67....... and 3, 10, 17........
For first A.P., first term, a = 63 and common
difference, d = 2
Then, its nth term = 63 + (n – 1) × 2 ...(i)
For second A.P., first term, a’ = 3 and common
difference, d’ = 7
Then, its nth term = 3 + (n − 1) × 7
According to the question:
63 + (n − 1) × 2 = 3 + (n − 1) × 7
(n − 10 × 5 = 60
n − 1 = 12
n = 13
The 13^th term of both given APs are equal.

QQuestionuestion. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution. Let the required numbers in A.P. are (a – d), a, (a + d) respectively.
Now, a – d + a + a + d = 15
3a = 15 ⇒ a = 5
According to question, number is
100(a – d) + 10a + a + d = 111a – 99d
Number on reversing the digits is
100(a + d) + 10a + a – d = 111a + 99d
Now, as per given condition in question,
(111a – 99d) – (111a + 99d) = 594
⇒ –198d = 594 ⇒ d = –3
So, digits of number are [5 – (–3), 5, (5 + (–3)]
= 8, 5, 2
Required number is 111 × (5) – 99 × (– 3)
= 555 + 297 = 852
The number is 852.

Question. The sum of the first 5 terms of an AP and the sum of the first 7 terms of the same AP is 167. If the sum of the first 10 terms of this AP is 235, find the sum of its first 20 terms.
Solution. Let a be the first term, d be the common difference and n be the number of terms of an AP.
We know that
S_n =n/2[2a + (n – 1)d]
Sum of first five terms, S5
S₅ =5/2[2a + (5 – 1)d] =5/2[2a + 4d]
S₅ = 5[a + 2d]
S₅ = 5a + 10d ...(i)
Sum of first seven terms, S7
S₇ =7/2[2a + (7 – 1)d]
= 7/2[2a + 6d]
S₇ = 7[a + 3d]
S₇ = 7a + 21d ...(ii)
Now, by given condition
S₅ + S7 = 167
⇒ 5a + 10d + 7a + 21d = 167
[using equation (i) & equation (ii)]
⇒ 12a + 31d = 167 ...(iii)
Also, it is given that sum of first 10 terms of this AP is 235.
⇒ S₁₀ = 235
⇒10/2[2a + (10 – 1)d] = 235
⇒ 5[2a + 9d] = 235
⇒ 2a + 9d = 235/5 = 47
⇒ 2a + 9d = 47 ...(iv)
Multiplying equation (iv) by 6 and subtracting
it from equation (iii), we get
(12a + 31d) – (12a + 54d) = 167 – 282
⇒ 23d = 115 ⇒ d = 5
Putting the value of d in equation (4), we get
2a + 9d = 47
⇒ 2a + 9(5) = 47
⇒ 2a + 45 = 47
⇒ 2a = 2 ⇒ a = 1
Sum of first 20 terms of this AP,
S₂₀ =20/2[2a + (20 – 1)d]
= 10[2(1) + 19(5)]
= 10[2 + 95]
= 10 × 97
= 970
Hence, the required sum of first 20 terms is 970.

Question. Among the natural numbers 1 to 49, find a number x, such that the sum of numbers preceding it is equal to sum of numbers succeeding it.
Solution. Let, the number be x.
1, 2, 3, 4, ..., x – 1, x, x + 1, ... 49
1 + 2 + 3 + 4 + ... + x – 1 = x + 1 + ... + 49
S_{x – 1} = S₅₀ – S_x
⇒ ( x -1 )x /2 = 49 x 50/2 − (x+ 1)x/2
⇒ x² – x = 2450 – x² – x
⇒ x² = 1225
⇒ x = ±35 (–35 is not between 1 to 49 therefore rejected)
x = 35.

Question. Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12. 
Solution. Let, the first term of A.P. be 'a’ and its common difference be 'd’
Given, a₃ = 16
i.e. a + (3 – 1)d = 16 [∵ an = a + (n – 1)d]
⇒ a + 2d = 19 ...(i)
And a7 = a₅ + 12
⇒ a + (7 – 1)d = a + (5 – 1)d + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12
⇒ d = 2
If we put the value of ‘d' in eduation (i), we get;
a + 2 × 2 = 16
⇒ a = 16 – 4 = 12
First term of AP = 12
Second term of AP = 12 + 2 = 14
Third term of AP = 14 + 2 = 16 and so on.
Hence, the required A.P is 12, 14, 16, 18, ....

Question. The 26^th, 11^th and the last term of an AP are 0, 3 and -1/5 respectively. Find the common difference and the number of terms.
Solution. It is given that:
26^th term of AP, a26 = 0
11^th term of AP, a11 = 3
Last term, L =-1/5
Let the AP contain n term, last term (L) is the n^th term.
Let first term be a and common difference be d of an AP.
We know that
an = a + (n – 1)d
Also, a₂₆ = 0 [Given]
⇒ a₂₆ = a + (26 – 1)d
⇒ 0 = a + 25d
⇒ a + 25d = 0
Also, a11 = 3
⇒ a + (11 – 1) d = 3
⇒ a + 10d = 3
Last term L = -1/5
a + (n – 1) d = -1/5
Subtracting equation (ii) from equation (i), we get
⇒ (a + 25d) – (a + 10d) = 0 – 3
⇒ a + 25d – a – 10d = 0 – 3
⇒ 15d = –3
⇒ d = -3/15 = -1/15
Putting the value of d in equation (i)
⇒ a + 25(-1/5) = 0
⇒ a – 5 = 0
⇒ a = 5
Putting the value of a and d in equation (iii)
⇒ a + (n – 1) d = -1/5
⇒ 5 + (n – 1) (-1/5) =-1/5
⇒ 25 – (n – 1) = –1
⇒ 25 + 1 = (n – 1)
⇒ n = 27
Hence, the common difference =-1/5 and number of terms = 27.

Question. Solve : 1 + 4 + 7 + 10 + .... + x = 287.
Solution. In the given AP, a = 1 and d = 3.
Let AP contains ‘n’ terms. Then, an = x
⇒ a + (n – 1)d = x
⇒ 1 + 3(n – 1) = x
⇒ n = x + 2/3
Further, Sn =n/2[first term + last term]
⇒ 287 = x + 2/3 x 2 [1 + x]
⇒ (x + 1) (x + 2) = 1722
or x² + 3x – 1720 = 0
⇒ x² + 43x – 40x – 1720 = 0
⇒ x(x + 43) – 40(x + 43) = 0
⇒ (x + 43) (x – 40) = 0
⇒ x – 40 = 0 (∵ x + 43 ≠ 0)
⇒ x = 40.
Thus, x = 40.

Question. A thief runs with a uniform speed of 100 m/ minute. After one minute a policeman after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief? 
Solution. Let, the total time to catch the thief be ‘n’ minutes.
speed of thief = 100 m/min.
Total distance covered by the thief = (100 n)
Here, the speed of the policeman in the first minute is 100 m/min, in the 2nd minute is 110 m/min, in the 3rd minute is 120 m/min and so on.
Then, the speed forms an A.P. with a constant
increasing speed of 10 m/min. thus, the series is :
100, 110, 120, 130
As the policeman starts after a minute (n – 1) Total distance covered by the policeman
= 100 + 110 + 120 + ... (n – 1) terms
100n = n - 1/2 [200 + (n – 2) × 10
[∴sn = n/2 [2a + (n- 1)]d]
⇒ 200n = n – 1[200 + 10n – 20]
⇒ 200n = (n – 1) (180 + 10n)
⇒ 200n = 180n – 180 + 10n2 – 10n
⇒ 10n² – 30n – 180 = 0
⇒ n² – 3n – 18 = 0
⇒ (n – 6) (n + 3) = 0
⇒ n = 6 [∵ n = – 3, is not possible]
Hence, the police'man takes (n – 1) = 5 minutes to catch the theif.